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Q20E

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Found in: Page 734

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Evaluate $$\iiint_{\text{E}}{\sqrt{{{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}}}}\text{dV}$$, where $$E$$ is enclosed by the planes $${\rm{z = 0}}$$ and $${\rm{z = x + y + 5}}$$ and by the cylinders $${{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ = 4}}$$ and $${{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ = 9}}$$.

The $$\int_{\rm{0}}^{{\rm{2\pi }}} {} \int_{\rm{2}}^{\rm{3}} {\int_{\rm{0}}^{{\rm{rcos\theta + rsin\theta + 5}}} {{{\rm{r}}^{\rm{2}}}} } {\rm{cos\theta dzdrd\theta = }}\frac{{{\rm{65\pi }}}}{{\rm{4}}}$$.

See the step by step solution

## Step 1:Determine the integral's value.

First, swap everything into cylindrical coordinates. The area defined by these coordinates is bordered by the planes $${\rm{z = 0}}$$ and $${\rm{z = rcos\theta + rsin\theta + 5}}$$ is located between the two cylinders with radii of $${\rm{2}}$$ and $${\rm{3}}$$. The triple integral is written as:

\begin{aligned}\int_{\rm{0}}^{{\rm{2\pi }}} {} \int_{\rm{2}}^{\rm{3}} {\int_{\rm{0}}^{{\rm{rcos\theta + rsin\theta + 5}}} {{{\rm{r}}^{\rm{2}}}} } \rm cos\theta dzdrd\theta &= \int_{\rm{0}}^{{\rm{2\pi }}} {\int_{\rm{2}}^{\rm{3}} {{{\rm{r}}^{\rm{3}}}} } {\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{\theta + }}{{\rm{r}}^{\rm{3}}}{\rm{cos\theta sin\theta + 5}}{{\rm{r}}^{\rm{2}}}{\rm{cos\theta drd\theta }}\\\rm &= \int_{\rm{0}}^{{\rm{2\pi }}} {\frac{{{\rm{65 + 65cos2\theta }}}}{{\rm{8}}}} {\rm{ + }}\frac{{{\rm{65cos\theta sin\theta }}}}{{\rm{4}}}{\rm{ + }}\frac{{{\rm{95cos\theta }}}}{{\rm{3}}}{\rm{d\theta }}\\\rm &= \frac{{{\rm{65\pi }}}}{{\rm{4}}}\end{aligned}

## Step 2: Result.

$$\int_{\rm{0}}^{{\rm{2\pi }}} {} \int_{\rm{2}}^{\rm{3}} {\int_{\rm{0}}^{{\rm{rcos\theta + rsin\theta + 5}}} {{{\rm{r}}^{\rm{2}}}} } {\rm{cos\theta dzdrd\theta = }}\frac{{{\rm{65\pi }}}}{{\rm{4}}}$$

Therefore, the$$\int_{\rm{0}}^{{\rm{2\pi }}} {} \int_{\rm{2}}^{\rm{3}} {\int_{\rm{0}}^{{\rm{rcos\theta + rsin\theta + 5}}} {{{\rm{r}}^{\rm{2}}}} } {\rm{cos\theta dzdrd\theta = }}\frac{{{\rm{65\pi }}}}{{\rm{4}}}$$.